A 100.0-mL buffer solution is 0.175 M in HClO and 0.150 M in NaClO. Part A: What is the initial pH of this solution?Part B: What is the pH after addition of 150.0 mg of HBr?Part C: What is the pH after addition of 85.0 mg of NaOH?

Initial pH of this pH  = 7.453

pH after addition of 150.0 mg of HBr  =  7.35

pH after addition of 85.0 mg of NaOH 0.154  = 7.56

Explanation:

Since Ka value isn't given  

so we use Ka value of HClO (hypo chlorous acid) = 3 x 10⁻⁸

pKa = - logKa = 7.52

Part A

Using Henderson equation

pH = pka + log

pH = 7.52 + log

pH = 7.453

Part B

pH after addition of 150 mg of HBr

moles of HBr    

Moles of NaOCl in 100 ml buffer solution = = 0.015

Moles of HClO in 100 ml buffer solution = = 0.0175

Since H⁺ concentration furnished by HBr acid make a common ion effect . So the following reaction carried out

ClO⁻ + H⁺ → HClO

So the remaining concentration of ClO⁻ in solution = 0.015 - 0.000185

                                                                                     = 0.0132            

moles of HClO = 0.0175 + 0.00185

                         = 0.0194

Using Henderson equation pH = Pka + log

                                                    = 7.52 + log

                                                     = 7.35

Part C

         pH after addition of 85 mg of HBr

moles of NaOH    

So the remaining concentration of ClO⁻ in solution = 0.015 + 0.00213

                                                                                     = 0.0171            

Moles of concentration of ClO⁻ = 0.171(M)

moles of HClO = 0.0175 - 0.00213

                         = 0.0154

Moles in 100 ml Buffer = 0.154(M)

Using Henderson equation pH = Pka + log

                                                    = 7.52 + log

                                                     = 7.56

what is the average time it took the car to travel 0.25 meters?

⇒ 2.23 seconds

what is the average time it took the car to travel 0.50 meters?

⇒ 3.13 seconds

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol